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(2x+3)(2x+3)=24.5
We move all terms to the left:
(2x+3)(2x+3)-(24.5)=0
We add all the numbers together, and all the variables
(2x+3)(2x+3)-24.5=0
We multiply parentheses ..
(+4x^2+6x+6x+9)-24.5=0
We get rid of parentheses
4x^2+6x+6x+9-24.5=0
We add all the numbers together, and all the variables
4x^2+12x-15.5=0
a = 4; b = 12; c = -15.5;
Δ = b2-4ac
Δ = 122-4·4·(-15.5)
Δ = 392
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{392}=\sqrt{196*2}=\sqrt{196}*\sqrt{2}=14\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-14\sqrt{2}}{2*4}=\frac{-12-14\sqrt{2}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+14\sqrt{2}}{2*4}=\frac{-12+14\sqrt{2}}{8} $
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